JEE Main 2022 June 27 – Shift 1 Maths Question Paper with Solutions

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SECTION – A

Multiple Choice Questions: This section contains 20 multiple choice questions. Each question has 4 choices (1), (2), (3) and (4), out of which ONLY ONE is correct.

Choose the correct answer :

1. The area of the polygon, whose vertices are the non-real roots of the equation

Answer (A)

Sol.

Let z = x + iy

x – iy = i(x² – y² + 2xiy)

x – iy = i(x² – y²) – 2xy

∴ x = –2yx or x² – y² = –y

x = 0 or

Case-I

x = 0

–y² = –y

y = 0, 1

Case-II

⇒

Area of polygon

2. Let the system of linear equations x + 2y + z = 2, αx + 3y – z = α, –αx + y + 2z = –α be inconsistent. Then α is equal to :

Answer (D)

Sol. x + 2y + z = 2

αx + 3y – z = α

–αx + y + 2z = –α

= 7 + 2α

∴ For no solution

3. If

(A) x, y, zare in A.P.

(B) x, y, zare in G.P.

(C)

(D)

Answer (C)

Sol.

Now,

a, b, c→ AP

1 – a, 1 – b, 1 – c→ AP

x, y, z→ HP

∴ [tex]\frac{1}{x},\frac{1}{y},\frac{1}{z}\rightarrow\textup{AP}[/tex]

4. Let

(A) 10

(B) 8

(C) 7

(D) 5

Answer (B)

Sol.

= bxdy + cy dy + a dy = ax dx – by dx + a dx

= cy dy + a dy – ax dx – a dx + b(x dy + y dx) = 0

= ax² – cy² + 2ax – 2ay – 2bxy = k

Above equation is circle

⇒ a = –c and b = 0

ax² + ay² + 2ax – 2ay = k

⇒ x² + y² + 2x – 2y = λ

Passes through (2, 5)

4 + 25 + 4 – 10 = λ⇒λ = 23

Circle ≡x² + y² + 2x – 2y – 23 = 0

Centre (–1, 1)

Shortest distance of (11, 6)

= 13 – 5

= 8

5. Let a be an integer such that

(A) –6

(B) –2

(C) 2

(D) 6

Answer (A)

Sol.

For limit to exist

LHL = RHL

∴ _{a = -6}

6. The number of distinct real roots of x⁴ – 4x + 1 = 0 is :

(A) 4

(B) 2

(C) 1

(D) 0

Answer (B)

Sol.

ƒ(x) = x⁴ – 4x + 1 = 0

ƒ′(x) = 4x³ – 4

= 4(x – 1) (x² + 1 + x)

⇒ Two solution

7. The lengths of the sides of a triangle are 10 + x², 10 + x² and 20 – 2x². If for x = k, the area of the triangle is maximum, then 3k² is equal to :

(A) 5

(B) 8

(C) 10

(D) 12

Answer (C)

Sol.

Area

⇒ 10 – x² = 2x²

3x² = 10

x = k

3k² = 10

8. If

(A) x²y′′ + xy′ – 25y = 0

(B) x²y′′ – xy′ – 25y = 0

(C) x²y′′ – xy′+ 25y = 0

(D) x²y′′ + xy′+ 25y = 0

Answer (D)

Sol.

Differentatingon both side

Square on both side

Diff on both side

xy′ + y′′x² + 25y = 0

9. If

Answer (B)

Sol.

for x = 1

10. The value of the integral

(A) 5e²

(B) 3e^–2

(C) 4

(D) 6

Answer (D)

Sol.

= 4 + 2 = 6

11. If

(A) 2 + log₂ 3

(B) 2 + log₃ 2

(C) 2 – log₃ 2

(D) 2 – log₂ 3

Answer (D)

Sol.

= |(2^y – 1)(2^x – 1)| = c

∵ y(1) = 1

∴ c = 1

= |(2^y – 1)(2^x – 1)| = 1

For x = 2

|(2^y – 1)3| = 1

Taking log to base 2.

∴ y = 2 – log₂ 3

12. In an isosceles triangle ABC, the vertex A is (6, 1) and the equation of the base BC is 2x + y = 4. Let the point B lie on the line x + 3y = 7. If (α, β) is the centroid of ΔABC, then 15(α + β) is equal to :

(A) 39

(B) 41

(C) 51

(D) 63

Answer (C)

Sol.

B(1, 2)

Let C(k, 4 – 2k)

Now AB² = AC²

5² + (–1)² = (6 – k)² + (–3 + 2k)²

⇒ 5k² – 24k + 19 = 0

(5k – 19)(k – 1) = 0

Centroid (α, β)

Now 15(α + β)

13. Let the eccentricity of an ellipse

(A) 29

(B) 31

(C) 32

(D) 34

Answer (B)

Sol.

⇒

a²(1 – e²) = b²

15a² = 16b² ⇒

From (i)

a² + b² = 15 + 16 = 31

14. If two straight lines whose direction cosines are given by the relations l + m – n = 0, 3l² + m² + cnl = 0 are parallel, then the positive value of cis :

(A) 6

(B) 4

(C) 3

(D) 2

Answer (A)

Sol. l + m – n = 0 ⇒n = l + m

3l² + m² + cnl = 0

3l² + m² + cl(l + m) = 0

= (3 + c)l² + clm + m² = 0

∵ Lines are parallel

D = 0

c² – 4(3 + c) = 0

c² – 4c – 12 = 0

(c – 4)(c + 3) = 0

c = 4 (as c> 0)

15. Let

(A) 0

(B) 1

(C) 2

(D) 3

Answer (A)

Sol.

Now,

= 2 – 3 – 2 = 0

⇒ –3 = 0 (Not possible)

⇒ No possible value of

16. Five numbers, x₁, x₂, x₃, x₄, x₅ are randomly selected from the numbers 1, 2, 3,….., 18 and are arranged in the increasing order (x₁<x₂<x₃<x₄<x₅). The probability that x₂ = 7 and x₄ = 11 is:

(A)

(C)

Answer (C)

Sol. Total cases_=18C5

Favourable cases

17. Let X be a random variable having binomial distribution B(7, p). If P(X = 3) = 5P(X = 4), then the sum of the mean and the variance of X is:

(A)

(B)

(C)

(D)

Answer (C)

Sol. Given P(X = 3) = 5P(X = 4) and n = 7

⇒

⇒ q = 5p and also p + q = 1

⇒

_Mean

Mean + Variance

18. The value of

(A) –1

(B)

(C)

(D)

Answer (B)

Sol.

19.

(A)

(B)

(C)

(D)

Answer (A)

Sol.

20. The boolean expression (~(p ∧q)) ∨q is equivalent to:

(A) q→ (p ∧q)

(B) p→q

(C) p→ (p→q)

(D) p→ (p∨q)

Answer (D)

Sol. Making truth table

SECTION – B

Numerical Value Type Questions: This section contains 10 questions. In Section B, attempt any five questions out of 10. The answer to each question is a NUMERICAL VALUE. For each question, enter the correct numerical value (in decimal notation, truncated/rounded-off to the second decimal place; e.g. 06.25, 07.00, –00.33, –00.30, 30.27, –27.30) using the mouse and the on-screen virtual numeric keypad in the place designated to enter the answer.

1. Let ƒ :R→R be a function defined by

Answer (99)

Sol.

∴

i.e. f(x) + f(1 – x) = 2

∴

= 49 × 2 + 1 = 99

2. If the sum of all the roots of the equation

Answer (45)

Sol. Let e^x = t then equation reduces to

⇒ 2t³ – 22t² + 81t – 45 = 0 …(i)

if roots of

⇒ α₁ + α₂ + α₃ = ln 45 ⇒ p = 45

3. The positive value of the determinant of the matrix A, whose

Answer (14)

Sol.

∴

4. The number of ways, 16 identical cubes, of which 11 are blue and rest are red, can be placed in a row so that between any two red cubes there should be at least 2 blue cubes, is _________.

Answer (56)

Sol. First we arrange 5 red cubes in a row and assume x₁, x₂, x₃, x₄, x₅ and x₆ number of blue cubes between them

Here, x₁ + x₂ + x₃ + x₄ + x₅ + x₆ = 11

and x₂, x₃, x₄, x₅≥ 2

So x₁ + x₂ + x₃ + x₄ + x₅ + x₆ = 3

No. of solutions = ⁸C₅ = 56

5. If the coefficient of x¹⁰ in the binomial expansion of

Answer (5)

Sol.

for

⇒ 180 – 3r – 2r = 60

⇒ r = 24

∴ Coeff. of

as l and 5 are coprime

k = 3 + exponent of 5 in ⁶⁰C₂₄

= 3 + (12 + 2 – 4 – 0 – 7 – 1)

= 3 + 2 = 5

6. Let

Answer (6*)

Sol.

Required area (above x-axis)

and

∴

⇒ k = 6

* A₁ is

Which tends to infinity if not mentioned above x-axis

7. If the sum of the first ten terms of the series

Answer (276)

Sol.

⇒

∴ m + n = 276

8. A rectangle R with end points of one of its sides as (1, 2) and (3, 6) is inscribed in a circle. If the equation of a diameter of the circle is

2x – y + 4 = 0, then the area of R is ________.

Answer (16)

Sol.

As slope of line joining (1, 2) and (3, 6) is 2 given diameter is parallel to side

∴

and

Area

9. A circle of radius 2 unit passes through the vertex and the focus of the parabola y² = 2x and touches the parabola

Answer (63)

Sol.

Let the equation of circle be

⇒

Radius

⇒

⇒

∵ This circle and parabola

⇒

⇒

⇒

10. Let the mirror image of the point (a, b, c) with respect to the plane 3x – 4y + 12z + 19 = 0 be

(a – 6, β, γ). If a + b + c = 5, then 7β – 9γ is equal to __________.

Answer (137)

Sol.

⇒ β = 8 + b

⇒ γ = -24 + c

⇒ 3a – 4b + 12c = 150….(1)

a + b + c = 5

⇒ 3a + 3b + 3c = 15….(2)

Applying (1) – (2), we get;

-7b + 9c = 135

7b – 9c = -135

7β – 9γ = 7(8 + b) – 9 (–24 + c)

= 56 + 216 + 7b – 9c.

= 56 + 216 – 135 = 137.

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